First!

best solution I can come up with so far is to count the unique dividers for each number. If it an even number of dividers it will be out in the end, else it will be on. so, 1st is on, 2nd off, 3rd off, 4th on, 5th off and so on.

But kinda boring to do all the counting.

you're not far from the solution.
The only point now, is to find an "elegant" way to find a way which numbers have an even number of dividers... (obviously i didn't post this to do the counting, would be extremely boring ;-) )

msg me if you want/find it so that other can search by themselves!

10

it will follow a pattern like this. 1 on 2 off, then 1 on 4 off, then 1 on 6 off...

1*on + (2*n)off

the end result will be: (1 = on 0 = off)
1. 100 (3 total)
2. 10000 (8 total)
3. 1000000 (15 total)
4. 100000000 (24 total)
5. 10000000000 (35 total)
6. 1000000000000 (48 total)
7. 100000000000000 (63 total)
8. 10000000000000000 (80 total)
9. 1000000000000000000 (99 total)
10. 1 (100 total, done)

and then just count the number of lines
this works with whatever number u choose. if there were 61 switches there would be 7 lights on.

if you dont want to make a table like this you can use the algorithm:
number of light on = sqrt(1+number of switches)-1
(rounded off to the next number in line, 8.53 would be rounded off to 9)

83713 switches for example:
sqrt(1+83713)-1 = 288.3337...
number of lights on would be 289.

sry for the math:D

waouh didn't catch a thing! except the solution... which happens to be true indeed ;-)

i'll post my solution later

it's somewhat diffiult to understand, but after 2 minutes i got it.

hint: reading the 1s and 0s in a row gives the final result.

The answer is 10 lights and i can give you the correspnonding numbers of the lights that are left on.

for any number n = p1^e1*p2^e2*....pn^en, its prime decomposition

where, p1,p2,...pn are distinct primes. the total number of factors of this number is k=(e1+1)(e2+1)....(e3+1), this number k is odd if and only if all these prime exponents e1,e2,...en are even. i.e. n is a perfect square.

since here every i th bulb is toggled as many times as there are factors of i so it means all perfect square indexed bulbs will be toggled an odd number of times while rest will be toggled an even number of times. since all bulbs were initially turned off, it means only bulbs with perfect square index will be left on.

now number of perfect squares less than equal to any given N is floor(sqrt(N)) which is for N=100 floor(sqrt(100))=10

so only 10 bulbs will be left on and they are indexed

1,4,9,16,25,36,49,64,81,100

way 2 stereotype yourself m8

Ten, because everyone else said so.

10, because you said that everyone else said 10.

10, i just know it.

Ten is correct. Because I am a mathematician.

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And because only squares have an odd number of divisors, cf. butcher_kgp's post.

Ten is correct, but not because you are a mathematician, because you're agardanchair, but because today is tenth. also, the new southpark is out, and it heavily involves cartman.

Yeah the solution is 10 well done